The problem asks us to find which of the given binomials is a factor of the polynomial $x^3 - 6x^2 + 11x - 6$.

1. Problem Description

The problem asks us to find which of the given binomials is a factor of the polynomial $x^3 - 6x^2 + 11x - 6$.

2. Solution Steps

We can use the Factor Theorem to check each binomial. The Factor Theorem states that a polynomial $f(x)$ has a factor $(x-a)$ if and only if $f(a) = 0$.

a) Check if $x-1$ is a factor.

Let $x-1 = 0$, so $x = 1$.

Then we substitute $x = 1$ into the polynomial $x^3 - 6x^2 + 11x - 6$:

$f(1) = (1)^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 12 - 12 = 0$.

Since $f(1) = 0$, then $x-1$ is a factor of $x^3 - 6x^2 + 11x - 6$.

b) Check if $2x+3$ is a factor.

Let $2x+3 = 0$, so $x = -\frac{3}{2}$.

Then we substitute $x = -\frac{3}{2}$ into the polynomial $x^3 - 6x^2 + 11x - 6$:

$f(-\frac{3}{2}) = (-\frac{3}{2})^3 - 6(-\frac{3}{2})^2 + 11(-\frac{3}{2}) - 6 = -\frac{27}{8} - 6(\frac{9}{4}) - \frac{33}{2} - 6 = -\frac{27}{8} - \frac{54}{4} - \frac{33}{2} - 6 = -\frac{27}{8} - \frac{108}{8} - \frac{132}{8} - \frac{48}{8} = \frac{-27 - 108 - 132 - 48}{8} = \frac{-315}{8} \neq 0$.

So $2x+3$ is not a factor.

c) Check if $x+7$ is a factor.

Let $x+7 = 0$, so $x = -7$.

Then we substitute $x = -7$ into the polynomial $x^3 - 6x^2 + 11x - 6$:

$f(-7) = (-7)^3 - 6(-7)^2 + 11(-7) - 6 = -343 - 6(49) - 77 - 6 = -343 - 294 - 77 - 6 = -720 \neq 0$.

So $x+7$ is not a factor.

d) Check if $x+1$ is a factor.

Let $x+1 = 0$, so $x = -1$.

Then we substitute $x = -1$ into the polynomial $x^3 - 6x^2 + 11x - 6$:

$f(-1) = (-1)^3 - 6(-1)^2 + 11(-1) - 6 = -1 - 6 - 11 - 6 = -24 \neq 0$.

So $x+1$ is not a factor.

Since $f(1) = 0$, $x-1$ is a factor.

3. Final Answer

x-1